Question

Question: Let N be a subgroup of a group G and let $\mathit{a}\mathbf{\in }\mathit{G}$.

(b) Prove that is N isomorphic to ${\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{N}\mathit{a}$. [Hint: Define$\mathit{f}\mathbf{:}\mathit{N}\mathbf{\to }{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{N}\mathit{a}$ by$\mathit{f}\left(n\right)\mathbf{=}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{n}\mathit{a}$ ]

Short answer

It has been proved that N is isomorphic to$a^{-1}Na$.

Step-by-step solution

Step-By-Step SolutionStep 1: Define a map f

Let,$f\left(n\right)=f\left(m\right)$for $m,n\in N$

then ,$a^{-1}na=a^{-1}ma$

Multiplying by $a^{-1}$on the left, and on the right,

We get, $n=m$

Hence, fis injective.

Show that fis homomorphism

Let, $m,n\in N$

$$\begin{gathered} f\left(mn\right)=a^{-1}mna \\ =a^{-1}ma{a}^{-1}na \\ =\left(a^{-1}ma\right)\left(a^{-1}na\right) \\ =f\left(m\right)f\left(n\right) \end{gathered}$$

Hence, fis a homomorphism

Conclusion

Since fis a bijective homomorphism hence, it is an isomorphism.